# Question e32b6

Sep 15, 2015

I_(3(aq))^(-) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)

#### Explanation:

• Oxidation:

2S_2O_(3(aq))^(2–) -> S_4O_(6(aq))^(2–) + 2e^-

• Reduction: I_(3(aq))^(–) + 2e^(-) -> 3 I_((aq))^(–)

the overall reaction then:

• Redox:

I_(3(aq))^(–) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)

From the balanced equation, we can say that:

(1)/"1" * n_(I_3^(–)) = (1)/"2" * n_(S_2O_3^(2–)).

where the denominators 1 and 2 were taking from the corresponding coefficients of I_3^– and S_2O_3^(2–) respectively.

Now,

[I_3^–] . V = (1)/"2" * [S_2O_3^(2–)] * V^'

(since $n = C \cdot V$), where

${V}^{'} = \text{10.50 mL }$ and $\text{ "V = "15.00 mL}$

Therefore,

[I_3^–] = (1)/"2" * [S_2O_3^(2–)] * (V')/"V"

[I_3^–] = (1)/"2" * "0.0500M" * (10.50 cancel("mL"))/(15.00 cancel("mL")) = "0.0175M"#

(rounded to 3 significant figures)

I would like to suggest the following video, that deals with a similar example in terms of stoichiometry of a redox reaction.

Sep 15, 2015

$\left(a\right) \text{ } {I}_{3 \left(a q\right)}^{-} + 2 {S}_{2} {O}_{3 \left(a q\right)}^{2 -} \rightarrow {S}_{4} {O}_{6 \left(a q\right)}^{2 -} + 3 {I}_{\left(a q\right)}^{-}$

$\left(b\right) \text{ "0.0175"mol/l}$

#### Explanation:

Thiosulfate ions give out electrons:

$2 {S}_{2} {O}_{3 \left(a q\right)}^{2 -} \rightarrow {S}_{4} {O}_{6 \left(a q\right)}^{2 -} + 2 e$

These are taken in by ${I}_{3}^{-}$ which are reduced:

${I}_{3 \left(a q\right)}^{-} + 2 e \rightarrow 3 {I}_{\left(a q\right)}^{-}$

${I}_{3 \left(a q\right)}^{-} + 2 {S}_{2} {O}_{3 \left(a q\right)}^{2 -} \rightarrow {S}_{4} {O}_{6 \left(a q\right)}^{2 -} + 3 {I}_{\left(a q\right)}^{-}$

(b).

$c = \frac{n}{v}$

$n = c \times v$

So no. moles ${S}_{2} {O}_{3}^{2 -} = 0.05 \times 0.0105 = 5.25 \times {10}^{- 4}$

From the equation we can see that the number of moles of ${I}_{3}^{-}$ must be half of this:

no. moles ${I}_{3}^{-} = \frac{5.25 \times {10}^{- 4}}{2} = 2.625 \times {10}^{- 4}$

$c = \frac{n}{v}$

So $\left[{I}_{3 \left(a q\right)}^{-}\right] = \frac{2.625 \times {10}^{- 4}}{0.015} = 0.0175 \text{mol/l}$